Largest Rectangle in Histogram

Given n non-negative integers representing the histogram's bar height where the width of each bar is 1, find the area of largest rectangle in the histogram.

Above is a histogram where width of each bar is 1, given height = [2,1,5,6,2,3].

 

The largest rectangle is shown in the shaded area, which has area = 10 unit.

 

For example,
Given heights = [2,1,5,6,2,3],
return 10.

思路1：brute force 暴露破解法。（注意数值可能为0的情况）。

　　　　遍历数组每一个元素，以改元素为起点向右遍历，求以该元素为起点的面积最大值。

public class Solution {
    /**
     * @param height: A list of integer
     * @return: The area of largest rectangle in the histogram
     */
    public int largestRectangleArea(int[] height) {
        if (height == null || height.length == 0) {
            return 0;
        }
        int maxArea = 0;
        for (int i = 0; i < height.length; i++) {
            int min = height[i];
            maxArea = Math.max(maxArea, min);
            for (int j = i + 1; j < height.length && height[j] != 0; j++) {
                min = Math.min(min, height[j]);
                maxArea = Math.max(maxArea, min * (j - i + 1));
            }
        }
        return maxArea;
    }
}

 Evolution ： 可以发现如果height[i] < height[i - 1] ， 那么 以height[i] 为起点的面积一定小于以height[i - 1] 为起点的面积，因此可以进行减枝操作。

public class Solution {
    public int largestRectangleArea(int[] heights) {
        if (heights == null || heights.length == 0) {
            return 0;
        }
        int maxArea = 0;
        for (int i = 0; i < heights.length; i++) {
            if (i > 0 && heights[i] < heights[i - 1]) {
                continue;
            }
            int min = heights[i];
            maxArea = Math.max(maxArea, min);
            for (int j = i + 1; j < heights.length && heights[j] != 0; j++) {
                min = Math.min(min, heights[j]);
                maxArea = Math.max(maxArea, min * (j - i + 1));
            }
        }
        return maxArea;
    }
}